How do I solve (e^x)-xe^x less than or equal to 0, using a number line? do i factor first?

Question

cans someone factor this first then show me how to use it with number line

Answer 1

factor to: (e^x)(1-x)<=0

ln both sides: (xlne)(1-x)<=ln0

simplify: x-x^2<=1

simple quadratic x^2-x+1=0

Answer 2

Factoring would help to simplify it first...

(e ^ x) - x e^x  Ã¢Â‰Â¤ 0

Factor out e^x

e^x ( 1 - x) ≤ 0

This tells us that 1 is a special point on the numberline.

We know this because it creates a multiply by 0, and anything times 0 is 0.

Anyhow now apply the natural log to both sides

ln ( e^x  ) * ( 1 - x)  Ã¢Â‰Â¤ ln (0)

Remember ln e = 1

x * ( 1 - x)  Ã¢Â‰Â¤ 1

Now distribute

x - x² ≤ 1

Get everything to one side and you have a quadtractic...

-x² + x  - 1 ≤ 0

Distribute the negative

x² - x  + 1 ≤ 0

Use the quadractic equation to determine the roots which are will be the important points on the number line.

For

a = 1, b = -1, c = 1

x =

-b ± √ ( b² - 4 a c)

----------------------

2 a

1  Ã‚± √ ( 1 - 4 * 1 * 1)

---------------------------

2

1  Ã‚± √ ( 1 - 4 )

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2

1  Ã‚± √ (-3)

----------------

2

1  Ã‚± 3i

------------

2

The roots are imaginary...

Answer 3

e^x (1-x)<=0, therefore

x>=1